Radius of Convergence

May 2010
98
16
Hello i need to determine the radius of convergense for the following series:

\(\displaystyle \sum_{n=0}^{\infty}\frac{1}{\sqrt{n!}}x^n\)

Im thinking the ratio test, but as im pretty uncertain about these types of problems, i'd like a second opinion, and a bit of help:)
 

HallsofIvy

MHF Helper
Apr 2005
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Yes, for any "radius of convergence" problem, the best way to go is either the ratio test or the root test (and most of the time the ratio test is much simpler than the root test).

Here, \(\displaystyle a_n= \frac{1}{\sqrt{n!}}x^n\) so \(\displaystyle a_{(n+1)}= \frac{1}{\sqrt{n+1}}x^{n+1}\).

Then \(\displaystyle \left|\frac{a_{n}}{a_{n+1}}\right|= \frac{\sqrt{n!}}{\sqrt{(n+1)!}}|x|\)\(\displaystyle = \sqrt{\frac{n!}{(n+ 1)!}}|x|\)

What is the limit of that as n goes to infinity?
For what x is that limit less than 1?
 
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May 2010
98
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The limit is 1, so x < 1 will make the limit less than 1 making the radius of convergence = 1 ?:)
 

skeeter

MHF Helper
Jun 2008
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The limit is 1, so x < 1 will make the limit less than 1 making the radius of convergence = 1 ?:)
note that \(\displaystyle \frac{n!}{(n+1)!} = \frac{1}{n+1}\) ... care to rethink your response?
 
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May 2010
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note that \(\displaystyle \frac{n!}{(n+1)!} = \frac{1}{n+1}\) ... care to rethink your response?
hmm, \(\displaystyle \frac{1}{n+1}\) diverges right?.. Im lost
 

Prove It

MHF Helper
Aug 2008
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hmm, \(\displaystyle \frac{1}{n+1}\) diverges right?.. Im lost
\(\displaystyle \frac{1}{n + 1} \to \frac{1}{\infty} \to 0\) as \(\displaystyle n \to \infty\).


So what do you think the radius of convergence has to be?
 
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May 2010
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Aha! So the radius of convergence is infinity or R? since the series goes to 0.
 

Prove It

MHF Helper
Aug 2008
12,883
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Aha! So the radius of convergence is infinity or R? since the series goes to 0.
Yes, since this limit is \(\displaystyle 0\), it is always \(\displaystyle <1\).

So the series converges for all \(\displaystyle x\).
 
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May 2010
98
16
Thanx a bunch everyone, im really struggling with these types of problems, but its getting better and better :)